Question

the expression $a(t)=1200cdot(1.07)^{t}$ models the amount of money in a certain savings account after $t$ years, where 1200 is the initial deposit and $7%$ is the annual interest rate. use the properties of exponents to rewrite the expression to determine the approximate monthly interest rate. move one answer to each box to complete the sentences. first, rewrite the expression as $1200cdot(1.07)^{t}$ using the identity property of multiplication. then, change the expression to $1200cdot(1.005654)^{12t}$ using the power of powers rule. simplify the expression to get $1.005654%$. this means that the monthly rate is

Explanation:

Step1: Apply the identity property of multiplication

The identity property of multiplication states that \(a = a^1\). So, \(1.07=1.07^{1}\), and the expression \(A(t)=1200\cdot(1.07)^{t}\) can be rewritten as \(A(t)=1200\cdot(1.07^{\frac{1}{12}\times12})^{t}\). Using the power - of - a - power rule \((a^{m})^{n}=a^{mn}\), we have \(A(t)=1200\cdot(1.07^{\frac{1}{12}})^{12t}\).

Step2: Calculate the monthly growth factor

We know that \(1.07^{\frac{1}{12}}\approx1.005654\). So the expression \(A(t)=1200\cdot(1.07^{\frac{1}{12}})^{12t}\) can be written as \(A(t)=1200\cdot(1.005654)^{12t}\). If we consider the general form of an exponential function \(A = P(1 + r)^{n}\), where \(P\) is the principal, \(r\) is the interest rate per period, and \(n\) is the number of periods. Here, for the monthly compounding case, if \(n = 12t\) (number of months) and \(A=1200\cdot(1.005654)^{12t}\), the growth factor per month is \(1.005654\). The monthly interest rate \(r\) is found by \(1 + r=1.005654\), so \(r = 0.005654\) or \(0.5654\%\)

Answer:

First, rewrite the expression as \(1200\cdot(1.07^{\frac{1}{12}})^{12t}\) using the identity property of multiplication. Then, change the expression to \(1200\cdot(1.005654)^{12t}\) using the power of powers rule. Simplify the expression to get a monthly rate of approximately \(0.5654\%\)